Question
Determine the $AP$ whose fourth term is $18$ and the difference of the ninth term from the fifteenth term is $30$ .

Answer

Given that, fourth term $a_4=18$ and the difference of the ninth term from the fifteenth term is $30 ,$
$a_{15}-a_9=30$
Clearly,
$a_4=18$
$\Rightarrow a+3 d=18$
$a_{15}-a_9=30$
$a+14 d-(a+8 d)=30$
$a+14 d-a-8 d=30$
$\Rightarrow 6 d=30$
$\Rightarrow d=\frac{30}{6}=5$
Substitute the value of $d$ in equation $(i)$
$a+3(5)=18$
$a=18-15=3$
First term $a_1=3$
Second term $a_2=a_1+d=3+5=8$
Third term $a_3=a_1+2 d=3+2(5)=13$
Therefore, the $AP$ is $3,8,13, \ldots$

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